a) x² - 5x + 6

= x² - 2x - 3x + 6

=(x² - 2x) - (3x + 6)

=x.(x - 2) - 3.(x - 2)

=(x-2).(x-3)

b) 3x²+9x-30

=3x²+15x-6x-30

=(3x²+15x) - (6x+30)

= 3x(x+5) - 6(x+5)

=(x+5).(3x-6)

c) x²-3x+2

=x²-2x-x+2

=(x²-2x) - (x-2)

=x(x-2)-(x-2)

=(x-2)(x-1)

a)x² - 5x + 6

= x² - 2x - 3x + 6

=x.(x - 2) - 3(x - 2)

=(x - 2).(x - 3)

b)3x² +9x -30

=3x² +15x - 6x -30

=3x.(x+5) - 6.(x + 5)

=(x+5).(3x - 6)

c)x² - 3x +2

=x² - 2x - x +2

=x.(x- 2) - 1.(x-2)

=(x-2).(x - 1)

d)x² - 9x +18

=x² - 6x -3x +18

=x.(x - 6) -3.(x - 6)

=(x - 6).(x - 3)

e)x² - 6x +8

=x² - 2x - 4x +8

=x.(x - 2)- 4.(x - 2)

=(x - 2).(x - 4)

f)x² - 5x -14

=x² + 2x - 7x - 14

=x.(x + 2) -7.(x + 2)

=(x + 2).(x - 7)

\(A=x^2+20y^2+8xy-4y+2015\)

\(=\left(x^2+8xy+16y^2\right)+\left(4y^2-4y+1\right)+2014\)

\(=\left(x+4y\right)^2+\left(2y-1\right)^2+2014\ge2014\forall x\)

Dấu "=" xảy ra khi:

\(\hept{\begin{cases}x+4y=0\\2y-1=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-2\\y=\frac{1}{2}\end{cases}}\)

Vậy GTNN của A là 2014 khi \(x=-2,y=\frac{1}{2}\)

\(B=\frac{x^2-2x+2016}{x^2}\)

\(=\frac{2016x^2-2.x.2016+2016^2}{2016x^2}\)

\(=\frac{\left(x^2-2.x.2016+2016^2\right)+2015x^2}{2016x^2}\)

\(=\frac{\left(x-2016\right)^2+2015x^2}{2016x^2}=\frac{\left(x-2016\right)^2}{2016x^2}+\frac{2015}{2016}\ge\frac{2015}{2016}\forall x\)

Dấu "=" xảy ra khi: \(x-2016=0\Rightarrow x=2016\)

Vậy GTNN của B là \(\frac{2015}{2016}\)khi x = 2016

(x-y+4)² - (2x+y-1)²

= (x-y+4 + 2x+y-1 ) . (x-y+4 - 2x+y-1)

1. \(4x^2-17xy+13y^2=4x^2-4xy-13xy+13y^2=4x\left(x-y\right)-13y\left(x-y\right)=\left(x-y\right)\left(4x-13y\right)\)

2. \(2x\left(x-5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

3. \(A=\left(2a-3b\right)^2+2\left(2a-3b\right)\left(3a-2b\right)+\left(2b-3a\right)^2\)

\(\Leftrightarrow\left(2a-3b\right)^2-2\left(2a-3b\right)\left(2b-3a\right)+\left(2b-3a\right)^2=\left(2a-3b-2b+3a\right)^2=\left(5a-5b\right)^2\)

\(=25\left(a-b\right)^2=25\cdot100=2500\)

\(a,ax+by+ay+bx=\left(ax+ay\right)+\left(by+bx\right)=a\left(x+y\right)+b\left(x+y\right)=\left(a+b\right)\left(x+y\right)\)

\(b,x^2y+xy+x+1=xy\left(x+1\right)+\left(x+1\right)=\left(xy+1\right)\left(x+1\right)\)

\(c,x^2-ax-bx+ab=x\left(x-a\right)-b\left(x-a\right)=\left(x-b\right)\left(x-2\right)\)

\(d,x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(xy-1\right)\left(x+y\right)\)

\(e,a\left(x^2+y\right)-b\left(x^2+y\right)=\left(a-b\right)\left(x^2+y\right)\)

\(f,x\left(a-2\right)-a\left(a-2\right)=\left(x-a\right)\left(a-2\right)\)